package leetcode._027;

/**
 * Given an array nums and a value val, remove all instances of that value in-place and return the new length.
 *
 * Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
 *
 * The order of elements can be changed. It doesn't matter what you leave beyond the new length.
 *
 * Example 1:
 *
 * Given nums = [3,2,2,3], val = 3,
 *
 * Your function should return length = 2, with the first two elements of nums being 2.
 *
 * It doesn't matter what you leave beyond the returned length.
 * Example 2:
 *
 * Given nums = [0,1,2,2,3,0,4,2], val = 2,
 *
 * Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
 *
 * Note that the order of those five elements can be arbitrary.
 *
 * It doesn't matter what values are set beyond the returned length.
 * Clarification:
 *
 * Confused why the returned value is an integer but your answer is an array?
 *
 * Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
 *
 * Internally you can think of this:
 *
 * // nums is passed in by reference. (i.e., without making a copy)
 * int len = removeElement(nums, val);
 *
 * // any modification to nums in your function would be known by the caller.
 * // using the length returned by your function, it prints the first len elements.
 * for (int i = 0; i < len; i++) {
 *     print(nums[i]);
 * }
 */

/**
 *
 * @author chengfangming
 * 中心思想就是把需要删除的元素与数组中的最后一个替换
 * 索引值i重置减1，相当于把最后一个元素与需要删除的元素再比较一次。
 * 数组长度减1
 */
public class Solution {
    public int removeElement(int[] nums, int val) {
        int length = nums.length;
        for (int i = 0; i < length; i++) {
            if (nums[i] == val) {
                nums[i] = nums[length - 1];
                i --;
                length--;
            }
        }
        return length;
    }
}
